Question

For an object placed at a distance $24\, m$ from a lens, a sharp focused image is observed on a screen placed at a distance $12\, cm$ from the lens A glass plate of refractive index $15$ and thickness $1 \,cm$ is introduced between lens and screen such that the glass plate plane faces parallel to the screen By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

• A. 0.8 m
• B. 3.2 m
• C. 1.2 m
• D. 5.6 m

Answer 1

The Correct Option is B

To solve this problem, we need to account for the effect of the glass plate on the formation of the image by the lens. We will use the lens formula and the concept of optical path length to determine the required shift in object distance.

1. The lens formula is given by: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) where \(f\) is the focal length of the lens, \(v\) is the image distance, and \(u\) is the object distance.
2. Initially, the object distance \(u = -24\, \text{cm}\) (negative, as per sign convention) and the image distance \(v = 12\, \text{cm}\).
3. Using the lens formula to find the focal length: \(\frac{1}{f} = \frac{1}{12} + \frac{1}{24}\) Solving, we find: \(f = 8 \, \text{cm}\)
4. When the glass plate is introduced, the effective optical path difference will change. The optical path length introduced by the plate is given by: \((\mu - 1) \cdot t\), where \(\mu = 1.5\) is the refractive index of the glass and \(t = 1\, \text{cm}\) is its thickness.
5. Thus, the additional optical path length is: \((1.5 - 1) \times 1 = 0.5 \, \text{cm}\).
6. This changes the effective image distance, thus: \(v_{\text{new}} = 12 - 0.5 = 11.5\, \text{cm}\)
7. We need to find the new object distance using the lens formula with the new image distance: \(\frac{1}{f} = \frac{1}{11.5} + \frac{1}{u_{\text{new}}}\)
8. Solving this equation for \(u_{\text{new}}\), we get: \(u_{\text{new}} \approx -20.8\, \text{cm}\)
9. The shift required in the object distance is: \(|24 - 20.8| = 3.2\, \text{cm}\),

Therefore, the object should be shifted by 3.2 cm to obtain a sharp image again on the screen.